Class 12 Mathematics Sample Paper With Solutions Set I

Please refer to Class 12 Mathematics Sample Paper With Solutions Set I below. These Class 12 Mathematics Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Mathematics Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all CBSE Sample Papers for Mathematics in Standard 12.

Sample Paper Class 12 Mathematics With Solutions Set I

SECTION – A

1. If a̅ and b̅ are such that |a̅| = 1/2,|b̅| = 4√3 and |a̅ x b̅| = 1/√3 , then find |a̅.b̅| .
Solution:

Class 12 Mathematics Sample Paper With Solutions Set I

2. If a̅ and b̅ are unit vectors, then what is the angle between a̅ and b̅ for a̅ -√2 b̅ to be a unit vector?
Solution: Let θ be the angle between a̅ and b̅

3. Find the distance between the planes r̅.(2î – 3ĵ + 4k̂) – 4 = 0 and r̅.(6î – 9ĵ + 18k̂) + 30 = 0 .
Solution: Note that the normals to the given planes are proportional so, these planes are parallel.
Now, r̅.(2î – 3ĵ + 4k̂) – 4 = 0 and r̅.(6î – 9ĵ + 18k̂) + 30 = 0

4. If A is a square matrix such that | A | = 5 , write the value of |AAT| .
Solution: As | AA^T | = | A|| A^T | = |A ||A | = 5X5 = 25.

5. If

Solution:

6. If

find the values of k and a.
Solution:

By equality of matrices, 3k = 4a, 2k = -8, – 5k = 5b
Solving these equation simultaneously we get :
k = -4, a = -3.

SECTION – B

7. Differentiate (sin 2x)x + sin^-1√3x with respect to x.
Solution:

Differentiate tan^-1(√1+x² – √1-x²/1+x² + √1-x² with respect to cos^-1 x² .
Solution:

8. Find k, if f(x) =

is continuous at x = 0 .
Solution:

By using (i), we get : k =1/2.

9. Find equation of normal to the curve ay² = x³ at the point where x coordinate is am2 .
Solution: Since x coordinate is am² so, ay² = (am² )³ = a³m⁶ ⇒ y = ± am³
Therefore the point is P(am² ,am³ ) and Q(am² ,-am³) .
Now differentiating the given curve w.r.t. x we get : 2ay dy/dx = 3x² ⇒ dy/dx = 3x²/2ay

10. Find :

Solution:

11. Find : ∫[log(logx) + 1/(log x)²] dx
Solution:

12. Evaluate :

Solution:

Evaluate :

Solution:

13. Solve the differential equation : (x+1) dy/dx – y = e^3x(x+1)3 .
Solution:

14. Solve the differential equation : 2ye^x/y dx + (y-2xe^x/y) dy = 0 .
Solution:

15. Ishan wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50 m, then its area will remain same, but if length is decreased by 10 m and breadth is decreased by 20 m, then its area will decrease by 5300 m². Using matrices, find the dimensions of the plot. Also give reason why he wants to donate the plot for a school.
Solution: Let the length and breadth of the of the land be x and y (in metres) respectively.
So, (x – 50)(y + 50) = xy ⇒ 50x – 50y = 2500 i.e., x – y = 50…(i)
And, (x -10)(y – 20) = xy – 5300 ⇒ -20x – 10y = -5500 i.e., 2x + y = 550…(ii)

Hence dimensions of the land are : length = 200 m and breadth =150 m.
Reason : Ishan knows the value of education, which may be the reason why he wishes to donate the plot for a school.

16. Prove that : 2sin^-1(3/5) – tan^-1(17/31) = π/4 .
Solution:

Solve the equation for x : cos(tan^-1 x) = sin(cot-1 3/4) .
Solution:

17. There are two bags A and B. Bag A contains 3 white and 4 red balls whereas bag B contains 4 white and 3 red balls. Three balls are drawn at random (without replacement) from one of the bags and are found to be two white and one red. Find the probability that these were drawn from bag B.
Solution: Let E₁ : selecting bag A, E₂ : selecting bag B and, A : getting 2 white and 1 red balls out of 3 drawn (without replacement)

18. Given that vectors a̅, b̅, c̅ form a triangle such that a̅ = b̅ + c̅ . Find p, q, r, s such that area of triangle is 5√6 where a̅ = pî + qĵ + rk̂, b̅ = sî + 3ĵ + 4k̂and c̅ = 3î + ĵ – 2k̂.
Solution: As it is given that a̅ = b̅ + c̅ ∴ pî + qĵ + rk̂= sî + 3ĵ + 4k̂+ 3î + ĵ – 2k̂
⇒ pî + qĵ + rk̂= (s+3) î + 4ĵ + 2k̂
On equating coefficients of î, ĵ, k̂on both sides, we get : p = s + 3,q = 4, r = 2 .

19. Find the equation of plane passing through the points A(3, 2, 1), B(4, 2, –2) and C(6, 5, –1) and hence find the value of λ for which A(3, 2, 1), B(4, 2, –2), C(6, 5, –1) and D( λ , 5, 5) are coplanar.
Solution: Equation of plane passing through the points A(3, 2, 1), B(4, 2, –2) and C(6, 5, –1) is :

⇒ 9(x – 3) – 7(y – 2) + 3(z -1) = 0 ⇒ 9x – 7y + 3z -16 = 0…(i) is the required plane.
Now if A(3, 2, 1), B(4, 2, –2), C(6, 5, –1) and D( λ , 5, 5) are coplanar then D must satisfy (i).
That is, 9λ – 7 x 5 + 3×5 -16 = 0 ⇒ λ = 4 .

Find the coordinates of the point where the line r̅ = (-î – 2ĵ – 3k̂) + λ (3î + 4ĵ + 3k̂) meets the plane which is perpendicular to the vector n̅ = î + ĵ + 3k̂and at a distance of 4√11 from origin.
Solution: Equation of plane perpendicular to n̅ = î + ĵ + 3k̂and at a distance of 4/√11 from origin is

SECTION – C

20. Let f : N → N be a function defined as f (x) = 4x² + 12x + 15 . Show that f : N→S is invertible (where S is the range of f). Find the inverse of f and hence find f ^-1(31) and f^-1(87) .
Solution: Here f : N → N, f (x) = 4x² + 12x +15
Let y be an arbitrary element of range S of function f. Then y = 4x² + 12x + 15, for some x in N, which implies that y = (2x + 3)2 + 6.
This gives x = √y-6-3/2 as y ≥ 6 .
Let us define g : S → N by g(y) = √y-6-3/2 .
Now, gof (x) = g = f (x)) = g(4x² + 12x +15) = g((2x + 3)² + 6) = √((2x+3)² + 6) -6 -3/2 = x .

21. Using properties of determinants, prove that :

Solution:

Using elementary row operations, find the inverse of the following matrix :

Solution:

Using elementary row operations, A = IA

22. Determine the intervals in which the function f (x) = x⁴ – 8x³ + 22x² – 24x + 21 is strictly increasing or strictly decreasing.
Solution: Given f (x) = x⁴ -8x³ + 22x² – 24x + 21 ⇒ f ‘(x) = 4x³ – 24x²+ 44x – 24 = 4(x -1)(x – 2)(x – 3)
For critical points, f ‘(x) = 4(x -1)(x – 2)(x -3) = 0 ∴ x =1,2,3

Find the maximum and minimum values of f (x) = sec x + log cos² x, 0 < x < 2π .
Solution: Here f (x) = sec x + log cos² x, 0 < x < 2π ⇒ f (x) = sec x + 2log cos x
Therefore f ‘(x) = sec x tan x – 2 tan x = tan x(sec x – 2)
For f ‘(x) = 0, tan x(sec x – 2) = 0 ⇒ tan x = 0, sec x – 2 = 0 ∴ x = π, π/3 , 5π/3 ∈ (0,2π)
Again f ”(x) = sec² x(sec x – 2) + tan² x sec x = sec x(sec² x + tan² x – 2sec x)
Note that f ”(π) = -1(1+ 0 + 2) = -3 < 0 so, x = π is a point of maxima,
f ”(π/3) = 2(4 + 3- 4) = 6 > 0 so, x = π/3 is a point of minima,
f ”(5π/3) = 2(4 + 3- 4) = 6 > 0 so, x = 5 π/3 is a point of minima.
Therefore maximum value of f (x) at x = π is f (π) = secπ + log cos² π = -1,
Minimum value of f (x) at x = π/3 is f(π/3) = sec (π/3) + log cos²(π/3) = 2 – log 4 or 2(1 – log 2) , Minimum value of f (x) at x = 5π/3 is
f(5π/3) = sec (5π/3) + log cos² (5π/3) = 2 – log 4 or 2(1- log 2) .

23. Using integration find the area of the region {(x, y): y²≤ 6ax and x² + y² ≤ 16a²}.
Solution:

24. Find the equation of the plane containing two parallel lines x-1/2 = y+1/-1 = z/3 and x/4 = y-2/-2 = z+1/6 . Also, find if the plane thus obtained contains the line x-2/3 = y-1/1 = z-2/5 or not.
Solution: Let L1 : x-1/2 = y+1/-1 = z/3 and L₂ : x/4 = y-2/-2 = z+1/6 The points in these lines are (1,–1, 0) and (0, 2, –1) respectively.
Let the d.r.’s of the normal to the required plane be A, B, C.
So equation of plane is, A(x -1) + B(y +1) + C(z – 0) = 0…(i)
As (0, 2, –1) lies on (i) so, -A + 3B – C = 0…(ii)
Also as the plane contains the lines so, the normal to the plane shall be ⊥ to these lines also.
That is, 2A – B + 3C = 0…(iii) (as d.r.’s of the line L1 are 2, –1, 3).
Solving (ii) and (iii), we get : A/8 = B/1 = C/-5 .
By (i), we have 8(x -1) +1(y +1) -5(z – 0) = 0 i.e., 8x + y – 5z = 7 .
Now let L₃ : x-2/3 = y-1/1 = z-2/5 . Clearly the point on it is M(2, 1, 2) and its d.r.’s are 3, 1, 5.
Since M satisfies the plane 8x + y – 5z = 7 as LHS : 8x + y – 5z = 8 x 2 + 1 -5 x 2 = 7 = RHS.
And 8x³+1×1+ (-5)x⁵ = 24 + 1- 25 = 0 i.e., normal of the plane is ⊥ to the line L₃ also.
Hence, the plane 8x + y – 5z = 7 contains the line L₃.

25. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available costing ₹ 5 per unit and ₹ 6 per unit respectively. One unit of food F1 contains 4 units of vitamin A and 3 units of minerals whereas one unit of food F2 contains 3 units of vitamin A and 6 units of minerals. Formulate this as a linear programming problem. Find the minimum cost of diet that consists of mixture of these two foods and also meets minimum nutritional requirement.
Solution: Let x and y (in units) of foods F₁ and F₂ be mixed.
To minimize : Z = (5x + 6y) in ₹
Subject to constraints :
x ≥ 0, y ≥ 0, 4x + 3y ≥ 80, 3x + 6y ≥ 100

Since feasible region is unbounded so, 124 may or may not be minimum value of Z.
To check, draw 5x + 6y < 124 .
As in the half plane 5x + 6y <124 , there is no point common with the feasible region.
Hence minimum value of Z is ₹ 124 for 12 units and 32/3 units of foods F1 and F2 respectively.

26. Three numbers are selected at random (without replacement) from first six positive integers. If X denotes the smallest of the three numbers obtained, find the probability distribution of X.
Also find the mean and variance of the distribution.
Solution: Total number of ways of selecting 3 numbers out of first 6 positive integers = ⁶C₃ = 20 .
As X : the smallest of the three numbers selected so, X = 1, 2, 3, 4.